From: ELDHO PAUL (cciein2006@gmail.com)
Date: Sat Jul 01 2006 - 10:14:04 ART
Aaron,
I've a clarification
>>>How do you get this answer.If you take the 7th bit in the 2nd octect.
There are four ones. So If we XOR four ones the answer should be zero.
Similarly the 6th bit. There are four ones. So the answer again should be
zero.
So the final answer becomes 00000000.00000000.00000000.00001000.
Please currect me if I'm wrong.
On 7/1/06, Aaron Pilcher <apilcher@itgcs.com> wrote:
>
> Here is what I came up with, logic included...(these do not take address
> overlapping into consideration)(2nd one, it may be too early for me it
> looks
> odd)
>
> Internetwork expert has a really good example of this on their site.
>
>
> 1st question
>
> ***************************************************
> xor 00000000.00000110.00000000.00001000
---------------------------------------------------
51.3.0.1 00110011.00000011.00000000.00000001
51.5.0.1 00110011.00000101.00000000.00000001
51.7.0.1 00110011.00000111.00000000.00000001
51.3.0.9 00110011.00000011.00000000.00001001
51.5.0.9 00110011.00000101.00000000.00001001
51.7.0.9 00110011.00000111.00000000.00001001
---------------------------------------------------
and 00110011.00000001.00000000.00000001
access-list 1 permit 51.1.0.1 0.6.0.8
***************************************************
2nd question
***************************************************
xor 01010110.10101100.00001011.00000000
---------------------------------------------------
192.172.8 11000000.10101100.00001000.00000000
192.172.9 11000000.10101100.00001001.00000000
192.172.10 11000000.10101100.00001010.00000000
192.172.11 11000000.10101100.00001011.00000000
150.16.1.0 10010110.00010000.00000001.00000000
---------------------------------------------------
and 10000000.00000000.00000000.00000000
access-list 1 permit 128.0.0.0 86.172.11.0
***************************************************
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of Sami
Sent: Saturday, July 01, 2006 7:04 AM
To: Faryar Zabihi (fzabihi)
Cc: ELDHO PAUL; ccielab@groupstudy.com
Subject: Re: Access list
All,
I am totally confused..which one is correct..everyone has different
answer..IE lab 20 section 9.1 solution guide says
51.3.0.1 0.0.0.8
51.5.0.1 0.2.0.8
Thanks
On 7/1/06, Faryar Zabihi (fzabihi) < fzabihi@cisco.com> wrote:
>
>
> What if I have,
> 192.172.1.0
> 192.172.2.0
> 192.172.3.0
> 192.172.4.0
> 192.172.5.0
> 192.172.6.0
> 192.172.7.0
> 192.172.8.0
> 192.172.9.0
> 192.172.10.0
> 192.172.11.0
> 192.172.12.0
> 192.172.13.0
> 192.172.14.0
> 192.172.15.0
> 150.16.1.0
>
> I need to keep
> 192.172.8.0
> 192.172.9.0
> 192.172.10.0
> 192.172.11.0
> 150.16.1.0
>
> Routing protocol RIP. Catch being ONLY ONE line ACL
> First I tried to grab the networks I need but the 150.16.1.0 I could not
> include with one line in an acl. Then I tried to get the ones I need to
> filter out and offset by 15. my question is how do I match
> 192.172.1.0
> 192.172.2.0
> 192.172.3.0
> 192.172.4.0
> 192.172.5.0
> 192.172.6.0
> 192.172.7.0
> 192.172.12.0
> 192.172.13.0
> 192.172.14.0
> 192.172.15.0
>
> Thanks in advance,
>
> Faryar
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
> ELDHO PAUL
> Sent: Saturday, July 01, 2006 6:20 AM
> To: Sami
> Cc: ccielab@groupstudy.com
> Subject: Re: Access list
>
> A more currect answer seems to be
> 51.1.0.1 0.0.0.8
> We will get the first portion 51.1.0.1 by doing an and opration of all
> the four ip addresses and the wildcard portion by doing the XOR
> operation of all the four ip addresses.
>
>
>
>
>
>
> On 7/1/06, ELDHO PAUL <cciein2006@gmail.com> wrote:
> >
> > I think we can match it with 2 access lists
> > 51.3.0.1 0.6.0.0
> > 51.3.0.9 0.6.0.0
> >
> >
> > On 7/1/06, Sami < sy1977@gmail.com> wrote:
> > >
> > > Group,
> > >
> > > One of the task say use minimum amount of line necessary to comple
> > > this task.
> > >
> > > 51.3.0.1
> > > 51.5.0.1
> > > 51.7.0.1
> > > 51.3.0.9
> > > 51.5.0.9
> > > 51.7.0.9
> > >
> > > How can I combine in one access list ?
> > >
> > > Thanks
> > >
> > > ____________________________________________________________________
> > > ___ Subscription information may be found at:
> > > http://www.groupstudy.com/list/CCIELab.html
>
> _______________________________________________________________________
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